/*
 * CS:APP Data Lab
 *
 * <Please put your name and userid here>
 *
 * bits.c - Source file with your solutions to the Lab.
 *          This is the file you will hand in to your instructor.
 *
 * WARNING: Do not include the <stdio.h> header; it confuses the dlc
 * compiler. You can still use printf for debugging without including
 * <stdio.h>, although you might get a compiler warning. In general,
 * it's not good practice to ignore compiler warnings, but in this
 * case it's OK.
 */

#if 0
/*
 * Instructions to Students:
 *
 * STEP 1: Read the following instructions carefully.
 */

You will provide your solution to the Data Lab by
editing the collection of functions in this source file.

INTEGER CODING RULES:
 
  Replace the "return" statement in each function with one
  or more lines of C code that implements the function. Your code 
  must conform to the following style:
 
  int Funct(arg1, arg2, ...) {
      /* brief description of how your implementation works */
      int var1 = Expr1;
      ...
      int varM = ExprM;

      varJ = ExprJ;
      ...
      varN = ExprN;
      return ExprR;
  }

  Each "Expr" is an expression using ONLY the following:
  1. Integer constants 0 through 255 (0xFF), inclusive. You are
      not allowed to use big constants such as 0xffffffff.
  2. Function arguments and local variables (no global variables).
  3. Unary integer operations ! ~
  4. Binary integer operations & ^ | + << >>
    
  Some of the problems restrict the set of allowed operators even further.
  Each "Expr" may consist of multiple operators. You are not restricted to
  one operator per line.

  You are expressly forbidden to:
  1. Use any control constructs such as if, do, while, for, switch, etc.
  2. Define or use any macros.
  3. Define any additional functions in this file.
  4. Call any functions.
  5. Use any other operations, such as &&, ||, -, or ?:
  6. Use any form of casting.
  7. Use any data type other than int.  This implies that you
     cannot use arrays, structs, or unions.

 
  You may assume that your machine:
  1. Uses 2s complement, 32-bit representations of integers.
  2. Performs right shifts arithmetically.
  3. Has unpredictable behavior when shifting if the shift amount
     is less than 0 or greater than 31.


EXAMPLES OF ACCEPTABLE CODING STYLE:
  /*
   * pow2plus1 - returns 2^x + 1, where 0 <= x <= 31
   */
  int pow2plus1(int x) {
     /* exploit ability of shifts to compute powers of 2 */
     return (1 << x) + 1;
  }

  /*
   * pow2plus4 - returns 2^x + 4, where 0 <= x <= 31
   */
  int pow2plus4(int x) {
     /* exploit ability of shifts to compute powers of 2 */
     int result = (1 << x);
     result += 4;
     return result;
  }

FLOATING POINT CODING RULES

For the problems that require you to implement floating-point operations,
the coding rules are less strict.  You are allowed to use looping and
conditional control.  You are allowed to use both ints and unsigneds.
You can use arbitrary integer and unsigned constants. You can use any arithmetic,
logical, or comparison operations on int or unsigned data.

You are expressly forbidden to:
  1. Define or use any macros.
  2. Define any additional functions in this file.
  3. Call any functions.
  4. Use any form of casting.
  5. Use any data type other than int or unsigned.  This means that you
     cannot use arrays, structs, or unions.
  6. Use any floating point data types, operations, or constants.


NOTES:
  1. Use the dlc (data lab checker) compiler (described in the handout) to 
     check the legality of your solutions.
  2. Each function has a maximum number of operations (integer, logical,
     or comparison) that you are allowed to use for your implementation
     of the function.  The max operator count is checked by dlc.
     Note that assignment ('=') is not counted; you may use as many of
     these as you want without penalty.
  3. Use the btest test harness to check your functions for correctness.
  4. Use the BDD checker to formally verify your functions
  5. The maximum number of ops for each function is given in the
     header comment for each function. If there are any inconsistencies 
     between the maximum ops in the writeup and in this file, consider
     this file the authoritative source.

/*
 * STEP 2: Modify the following functions according the coding rules.
 * 
 *   IMPORTANT. TO AVOID GRADING SURPRISES:
 *   1. Use the dlc compiler to check that your solutions conform
 *      to the coding rules.
 *   2. Use the BDD checker to formally verify that your solutions produce 
 *      the correct answers.
 */

#endif
// 1
/*
 * bitXor - x^y using only ~ and &
 *   Example: bitXor(4, 5) = 1
 *   Legal ops: ~ &
 *   Max ops: 14
 *   Rating: 1
 */
int bitXor(int x, int y) {  // 异或：两数bit不同时为1并且不同时为0
  return ~(~x & ~y) & ~(x & y);
  // (~x & ~y)  两数为0的bit此时为1，再取反意味着提取两数不同时为0的bit
  // 同理，~(x & y)是提取两数不同时为1的bit，再&就得到异或
}
/*
 * tmin - return minimum two's complement integer
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 4
 *   Rating: 1
 */
int tmin(void) { return 1 << 31; }
// 2
/*
 * isTmax - returns 1 if x is the maximum, two's complement number,
 *     and 0 otherwise
 *   Legal ops: ! ~ & ^ | +
 *   Max ops: 10
 *   Rating: 1
 */
int isTmax(int x) {
  // 原理：0x7fffffff 取反之后的结果 0x80000000 与 0 是唯二的 两个补码是自身的 数
  int negate1 = ~x;            // 取反
  int negate2 = ~negate1 + 1;  // 取取反结果的补码
  return (!(negate2 ^ negate1)) & !!negate1;
  // 取过补码之后，值是否变化  并排除0
}

/*
 * allOddBits - return 1 if all odd-numbered bits in word set to 1
 *   where bits are numbered from 0 (least significant) to 31 (most significant)
 *   Examples allOddBits(0xFFFFFFFD) = 0, allOddBits(0xAAAAAAAA) = 1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 2
 */
int allOddBits(int x) {
  int mask = 0xAA + (0xAA << 8);  // 0xAA == 0b10101010
  mask = mask + (mask << 16);     // 构造 0b10101010101010101010101010101010
  return !(mask ^ (mask & x));    // check 奇数位是否逐位相同
}
/*
 * negate - return -x
 *   Example: negate(1) = -1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 5
 *   Rating: 2
 */
int negate(int x) { return ~x + 1; }
// 3
/*
 * isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters '0'
 * to '9') Example: isAsciiDigit(0x35) = 1. isAsciiDigit(0x3a) = 0.
 *            isAsciiDigit(0x05) = 0.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 15
 *   Rating: 3
 */
int isAsciiDigit(int x) {
  int sign = 1 << 31;
  int upperBound = ~(sign | 0x39);  // 加上比0x39大的数后符号由正变负
  int lowerBound = ~0x30;           // 加上比0x30小的值时是负数
  upperBound = sign & (upperBound + x) >> 31;
  lowerBound = sign & (lowerBound + 1 + x) >> 31;
  return !(upperBound | lowerBound);
}
/*
 * conditional - same as x ? y : z
 *   Example: conditional(2,4,5) = 4
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 16
 *   Rating: 3
 */
int conditional(int x, int y, int z) {
  x = !!x;     // 变成bool
  x = ~x + 1;  // 1则变负
  return (x & y) | (~x & z);
}
/*
 * isLessOrEqual - if x <= y  then return 1, else return 0
 *   Example: isLessOrEqual(4,5) = 1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 24
 *   Rating: 3
 */
int isLessOrEqual(int x, int y) {
  int negX = ~x + 1;               // -x
  int addX = negX + y;             // y-x
  int checkSign = addX >> 31 & 1;  // y-x的符号
  int leftBit = 1 << 31;           // 最大位为1的32位有符号数
  int xLeft = x & leftBit;         // x的符号
  int yLeft = y & leftBit;         // y的符号
  int bitXor = xLeft ^ yLeft;  // x和y符号相同标志位，相同为0不同为1
  bitXor = (bitXor >> 31) & 1;  // 符号相同标志位格式化为0或1
  return ((!bitXor) & (!checkSign)) | (bitXor & (xLeft >> 31));
  // 返回1有两种情况：
  // 1. 符号相同 y-x 且 y-x>=0
  // 2. 符号不同 x<0
}
// 4
/*
 * logicalNeg - implement the ! operator, using all of
 *              the legal operators except !
 *   Examples: logicalNeg(3) = 0, logicalNeg(0) = 1
 *   Legal ops: ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 4
 */
int logicalNeg(int x) {
  // 只有0和INT_MIN的补码为本身，其余数为其相反数
  // INT_MIN的符号位为1，0的符号位为0
  // 负数移位 符号位不变是GCC的实现
  // 但负数移位是implementation-defined behaviour
  // 为了跨平台，特判掉负数
  return (x >> 31 & 1) | (((x | (~x + 1)) >> 31) + 1);
}
/* howManyBits - return the minimum number of bits required to represent x in
 *             two's complement
 *  Examples: howManyBits(12) = 5
 *            howManyBits(298) = 10
 *            howManyBits(-5) = 4
 *            howManyBits(0)  = 1
 *            howManyBits(-1) = 1
 *            howManyBits(0x80000000) = 32
 *  Legal ops: ! ~ & ^ | + << >>
 *  Max ops: 90
 *  Rating: 4
 */
int howManyBits(int x) {
  int b16, b8, b4, b2, b1, b0;
  int sign = x >> 31;
  x = (sign & ~x) | (~sign & x);
  // 如果x为正则不变，否则按位取反
  // （这样好找最高位为1的，原来是最高位为0的，这样也将符号位去掉了）

  // 不断缩小范围
  b16 = !!(x >> 16) << 4;  // 高十六位是否有1
  x = x >> b16;  // 如果有（至少需要16位），则将原数右移16位
  b8 = !!(x >> 8) << 3;  // 剩余位高8位是否有1
  x = x >> b8;  // 如果有（至少需要16+8=24位），则右移8位
  b4 = !!(x >> 4) << 2;  // 同理
  x = x >> b4;
  b2 = !!(x >> 2) << 1;
  x = x >> b2;
  b1 = !!(x >> 1);
  x = x >> b1;
  b0 = x;
  return b16 + b8 + b4 + b2 + b1 + b0 + 1;  // +1表示加上符号位
}
// float
/*
 * floatScale2 - Return bit-level equivalent of expression 2*f for
 *   floating point argument f.
 *   Both the argument and result are passed as unsigned int's, but
 *   they are to be interpreted as the bit-level representation of
 *   single-precision floating point values.
 *   When argument is NaN, return argument
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
unsigned floatScale2(unsigned uf) {
  int exp = (uf & 0x7f800000) >> 23;
  int sign = uf & (1 << 31);
  if (exp == 0) return uf << 1 | sign;
  if (exp == 255) return uf;
  exp++;
  if (exp == 255) return 0x7f800000 | sign;
  // 如果指数+1之后为指数为255则返回原符号无穷大
  return (exp << 23) | (uf & 0x807fffff);
  // 否则返回指数+1之后的原符号数
}
/*
 * floatFloat2Int - Return bit-level equivalent of expression (int) f
 *   for floating point argument f.
 *   Argument is passed as unsigned int, but
 *   it is to be interpreted as the bit-level representation of a
 *   single-precision floating point value.
 *   Anything out of range (including NaN and infinity) should return
 *   0x80000000u.
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
int floatFloat2Int(unsigned uf) {
  int sign = uf >> 31;
  int exp = ((uf & 0x7f800000) >> 23) - 127;
  int frac = (uf & 0x007fffff) | 0x00800000;
  if (!(uf & 0x7fffffff)) return 0;  // 如果原浮点值为0则返回0；
  if (exp > 31) return 0x80000000;
  // 如果真实指数大于31（frac部分是大于等于1的，1<<31位会覆盖符号位），返回规定的溢出值0x80000000u；
  if (exp < 0) return 0;  // 如果exp<0（1右移x位,x>0，结果为0）则返回0
  if (exp > 23)  // 首先把小数部分（23位）转化为整数（和23比较）
    frac <<= (exp - 23);
  else
    frac >>= (23 - exp);

  if (!((frac >> 31) ^ sign))  // 判断是否溢出：如果和原符号相同则直接返回
    return frac;
  else if (frac >> 31)
    // 否则如果结果为负（原来为正）则溢出返回越界指定值0x80000000u
    return 0x80000000;
  else
    return ~frac + 1;  // 否则原来为负，结果为正，则需要返回其补码（相反数）
}
/*
 * floatPower2 - Return bit-level equivalent of the expression 2.0^x
 *   (2.0 raised to the power x) for any 32-bit integer x.
 *
 *   The unsigned value that is returned should have the identical bit
 *   representation as the single-precision floating-point number 2.0^x.
 *   If the result is too small to be represented as a denorm, return
 *   0. If too large, return +INF.
 *
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. Also if, while
 *   Max ops: 30
 *   Rating: 4
 */
unsigned floatPower2(int x) {
  int INF = 0xff << 23;
  int exp = x + 127;
  if (exp < 0 && exp > -24) {  // 考虑denorm就可以 23 是frac部分的位数
    return 0x00040000 >> (~exp + 1);
  } else if (exp < -23) {
    return 0;
  }
  if (exp >= 255) {  // 如果exp大于等于255则为无穷大或越界了
    return INF;
  }
  return exp << 23;
}